Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

On 2025-02-07 23:13:04 +0000, olcott said:

Experts in the C programming language will know that DD
correctly simulated by HHH cannot possibly reach its own
"if" statement.

Wrong, they understand that nothing below exludes the possibility that
HHH is a program that can correctly simulate DD to its "if" statement.
The code of HHH might exlude that but that is not sohwn below.

The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH.

No, it does not. DD as quoted below pecifies nothing about the behaviour
of HHH, only its argument types and return type.

This makes HHH necessarily correct to reject its input as
non-halting.

No, it does not. It is not correct to reject the input to HHH as non-halting
unless it really is non-halting.

typedef void (*ptr)();
int HHH(ptr P);
 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 int main()
{
   HHH(DD);
}

https://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD

No, it has not. There is no DD there.

The halting problem has always been a mathematical mapping
from finite strings to behaviors.

The problem is not a mappling. It requires that its solution is.
--
Mikko